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When \( \theta \) is small, find an approximation for \( \cos 30^\circ + \theta \sin 2\theta \), giving your answer in the form \( a + b\theta^2 \). - AQA - A-Level Maths Pure - Question 3 - 2017 - Paper 1
Question 3
When \( \theta \) is small, find an approximation for \( \cos 30^\circ + \theta \sin 2\theta \), giving your answer in the form \( a + b\theta^2 \).
Worked Solution & Example Answer:When \( \theta \) is small, find an approximation for \( \cos 30^\circ + \theta \sin 2\theta \), giving your answer in the form \( a + b\theta^2 \). - AQA - A-Level Maths Pure - Question 3 - 2017 - Paper 1
Step 1
Use approximation for cosine and sine
Answer
For small values of ( \theta ), we can use the approximations: [ \cos x \approx 1 - \frac{x^2}{2} ] and [ \sin x \approx x ] Substituting in these approximations: [ \cos 30^\circ + \theta \sin 2\theta \approx \cos 30^\circ + \theta \cdot 2\theta = \cos 30^\circ + 2\theta^2 ]
Step 2
Substitute in the value for cos 30°
Answer
We know that:[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 1 - \frac{(30^\circ)^2}{2} \text{ (where } 30^\circ = \frac{\pi}{6} \text{ in radians)}] Therefore, substituting, we get: [ \cos 30^\circ \approx 1 - \frac{(\frac{\pi}{6})^2}{2} = 1 - \frac{\pi^2}{72}]
Step 3